(6x-4)(7x+1)=0

Solution for (6x-4)(7x+1)=0 equation:

(6x-4)(7x+1)=0

We multiply parentheses ..

(+42x^2+6x-28x-4)=0

We get rid of parentheses

42x^2+6x-28x-4=0

We add all the numbers together, and all the variables

42x^2-22x-4=0

a = 42; b = -22; c = -4;
Δ = b2-4ac
Δ = -222-4·42·(-4)
Δ = 1156
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1156}=34$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-34}{2*42}=\frac{-12}{84}
=-1/7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+34}{2*42}=\frac{56}{84}
=2/3
$